Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
double dp[3001][3001];
double solve(double a[],int i,int x){
	if(x==0)return 1;
  	if(i==0)return 0;
  	if(dp[i][x]>-0.9)return dp[i][x];
  	return dp[i][x]=a[i]*solve(a,i-1,x-1)+(1-a[i])*solve(a,i-1,x);
}


int main(){
	int n;
    cin>>n;
  	double a[n];
  	memset(dp,-1,sizeof(dp));
  	for(int i=1;i<=n;i++){
      cin>>a[i];
    }
  	cout<<fixed<<setprecision(10)<<solve(a,n,(n+1)/2);


	return 0;
}